FREE PROBABILITY OF TYPE B AND ASYMPTOTICS OF …Many B’s around: bi-free probability...

FREE PROBABILITY OF TYPE B AND

ASYMPTOTICS OF FINITE-RANKPERTURBATIONS OF RANDOM MATRICES

Dima Shlyakhtenko, UCLA

Free Probability and Large N Limit, V

Many B ’s around:

Many B ’s around:

◮ bi-free probability (Voiculescu’s talk)

Many B ’s around:


◮ B-valued probability (amalgamation over a subalgebra)

Many B ’s around:



◮ B-valued bi-free (Skoufranis’s talk)

Many B ’s around:



◮ B-valued bi-free (Skoufranis’s talk)

◮ type B free probability (this talk)

Wigner’s Semicircle LawLet A(N) be an N × N random matrix so that

{Re(Aij), Im(Aij) : 1 ≤ i < j ≤ N} ∪ {Akk : 1 ≤ k ≤ N}are iid real Gaussians of variance N−1/2(1 + δij). LetλA

1(N) ≤ · · · ≤ λA

N(N) be the eigenvalues of A(N), and let

µAN =

1

N

∑

j

δλAj(N).

Then as N → ∞

E[µAN ] → semicircle law =

1

π

√

2 − t2χ[−√

2,√

2]dt

Voiculescu’s Asymptotic FreenessA(N) as before, B(N) diagonal matrix with eigenvaluesλB

1(N) ≤ · · · ≤ λB

N(N). Assume that

µBN =

1

N

∑

j

δλBj(N) → µB .

Then A(N) and B(N) are asymptoically freely independent. Inparticular,

µA+BN → µA

⊞ µB .

Voiculescu’s Asymptotic FreenessA(N) as before, B(N) diagonal matrix with eigenvaluesλB

1(N) ≤ · · · ≤ λB

N(N). Assume that

µBN =

1

N

∑

j

δλBj(N) → µB .

Then A(N) and B(N) are asymptoically freely independent. Inparticular,

µA+BN → µA

⊞ µB .

Example: B = 3PN with PN a projection of rank N/2.

Analytic Subordination and Free Convolution[Biane,Voiculescu,...]

To compute η = µA ⊞ µB define Gν =´

1

z−tdν(t). Then there

exist analytic functions ωA, ωB : C+ → C+ uniquely determined by

◮ GµA(ωA(z)) = GµB (ωB(z)) = Gη(z)

◮ ωA(z) + ωB(z) = z + 1/Gη(z)

◮ limy↑∞ ωA(iy)/(iy) = limy→∞ ω′A(iy) = 1 and same for ωB .



1



◮ GµA(ωA(z)) = GµB (ωB(z)) = Gη(z)

◮ ωA(z) + ωB(z) = z + 1/Gη(z)


Put Fν(z) = 1/Gν(z). Then Rν(z) = F−1ν (z) + z so that

RµA(z) + RµB (z) = RµA⊞µB (z) becomes

F−1

µA (z) + F−1

µB (z) = z + F−1

µA⊞µB (z).



1



◮ GµA(ωA(z)) = GµB (ωB(z)) = Gη(z)⇔ FµA(ωA(z)) = FµB (ωB(z)) = Fη(z)

◮ ωA(z) + ωB(z) = z + 1/Gη(z)




F−1

µA (z) + F−1

µB (z) = z + F−1

µA⊞µB (z).

ωA = F−1

µA ◦ FµA⊞µB ωB = F−1

µB ◦ FµA⊞µB .



1



◮ GµA(ωA(z)) = GµB (ωB(z)) = Gη(z)⇔ FµA(ωA(z)) = FµB (ωB(z)) = Fη(z)

◮ ωA(z) + ωB(z) = z + 1/Gη(z)⇔ ωA(z) + ωB(z) = z + Fη(z)




F−1

µA (z) + F−1

µB (z) = z + F−1

µA⊞µB (z).

ωA = F−1

µA ◦ FµA⊞µB ωB = F−1

µB ◦ FµA⊞µB .

Finite-rank perturbations [Ben Arous, Baik, Peche].

Let A(N) be as before but consider B(N) a finite rank matrix (e.g.BN = θQN) with QN rank 1 projection.

Finite-rank perturbations [Ben Arous, Baik, Peche].

Let A(N) be as before but consider B(N) a finite rank matrix (e.g.BN = θQN) with QN rank 1 projection.Semicircular limit for A(N) + B(N) but there may or may not beoutlier eigenvalues:

θ = 3 θ = 0.5

Finite rank perturbations and freeness?

It was discovered (Capitaine, Belischi-Bercovici-Capitain-Fevrier)that the description of the outlier involves free subordinationfunctions. For example, if AN is GUE and BN has 1 eigenvalue θand the rest zero, then we set

(ωA, ωB) = subordination functions for η ⊞ δ0

with η = semicircle law, i.e., ωA(z) = F−1η (z), ωB(z) = z ,

then there will be an outlier at θ′ = ωA(θ) (i.e. Gµ(θ′) = 1/θ).

Finite rank perturbations and freeness?

It was discovered (Capitaine, Belischi-Bercovici-Capitain-Fevrier)that the description of the outlier involves free subordinationfunctions. For example, if AN is GUE and BN has 1 eigenvalue θand the rest zero, then we set

(ωA, ωB) = subordination functions for η ⊞ δ0

with η = semicircle law, i.e., ωA(z) = F−1η (z), ωB(z) = z ,

then there will be an outlier at θ′ = ωA(θ) (i.e. Gµ(θ′) = 1/θ).

Why?! Is there still some free independence involved?

Another look at laws of random matrices

We consider the 1/N expansion of the law of AN + BN :

µA+BN = µA+B +

1

Nµ̇A+B + o(N−1).

The idea is that moving 1 eigenvalue out of N gives a perturbationof µA+B which is of order 1/N. Our aim is to compute µ̇A+B .Thus we want to keep track of the pair µA+B , µ̇A+B and not justµA+B (ordinary free probability).

Infinitesimal free probability theory [Belinschi-D.S, 2012]

To encode such questions we consider an infinitesimal probabilityspace (A, φ, φ′) where A is a unital algebra, φ, φ′ : A → C are linearfunctionals and φ(1) = 1, φ′(1) = 0.

Example

Let (A, φt) be a family of probability spaces, and assume thatφt = φ+ tφ′ + o(t). Then (A, φ, φ′) is an infinitesimal probabilityspace.

Infinitesimal free probability theory [Belinschi-D.S, 2012]

To encode such questions we consider an infinitesimal probabilityspace (A, φ, φ′) where A is a unital algebra, φ, φ′ : A → C are linearfunctionals and φ(1) = 1, φ′(1) = 0.

Example

Let (A, φt) be a family of probability spaces, and assume thatφt = φ+ tφ′ + o(t). Then (A, φ, φ′) is an infinitesimal probabilityspace.Eg: Xt family of random variables and you define φt : C[t] → C byφt(p) = E(p(Xt)).

Infinitesimal freeness, ctd.

We say that A1,A2 ⊂ A are infinitesimally free if the freenesscondition in (A, φt = φ+ tφ′) holds to order o(t).

Infinitesimal freeness, ctd.

We say that A1,A2 ⊂ A are infinitesimally free if the freenesscondition in (A, φt = φ+ tφ′) holds to order o(t).In other words, the following conditions holds whenevera1, . . . , ar ∈ A are such that ak ∈ Aik , i1 6= i2, i2 6= i3, . . . andφ(a1) = φ(a2) = · · · = φ(an) = 0:

φ(a1 · · · ar ) = 0;

φ′(a1 · · · ar ) =

r∑

j=1

φ(a1 · · · aj−1φ′(aj)aj+1 · · · ar ).

Free probability of type B [Biane-Goodman-Nica, 2003]

We introduced infinitesimal free probability theory to get a betterunderstanding of type B free probability introduced byBiane-Goodman-Nica. Their motivation was purely combinatorial:free probability is obtained from classical probability by replacingthe lattice of all partitions by the lattice of (type A) non-crossingpartitions:

Free probability of type B [Biane-Goodman-Nica, 2003]

We introduced infinitesimal free probability theory to get a betterunderstanding of type B free probability introduced byBiane-Goodman-Nica. Their motivation was purely combinatorial:free probability is obtained from classical probability by replacingthe lattice of all partitions by the lattice of (type A) non-crossingpartitions:Non-crossing partition (of type A): partition of (1, . . . , n) sothat if i < j < k < l and i ∼ k , j ∼ l then i ∼ l .

1

•2

•3

•4

•5

•6

•

Non-crossing partitions of type B

Non-crossing partition (of type B): non-crossing partition of(1, 2, . . . , n,−1,−2, . . . ,−n) so that if B is a block then so is −B.Either π = π+ ⊔ π− with π+ partition of (1, . . . , n) or there is azero block B so that B = −B.

1

•2

•3

•4

•5

•6

•−1

•−2

•−3

•−4

•−5

•−6

•



1

•2

•3

•4

•5

•6

•−1

•−2

•−3

•−4

•−5

•−6

•

zero block



1

•2

•3

•4

•5

•6

•−1

•−2

•−3

•−4

•−5

•−6

•

zero block

Type A non-crossing partitions have to do with geodesics in(Sn, transpositions) connecting 1 and (1 . . . n).

π 7→∏

C block of π

(cyclic permutation of C )



1

•2

•3

•4

•5

•6

•−1

•−2

•−3

•−4

•−5

•−6

•

zero block

Type A non-crossing partitions have to do with geodesics in(Sn, transpositions) connecting 1 and (1 . . . n).

π 7→∏

C block of π

(cyclic permutation of C )

Type B: same for the hyperoctahedral group.

Type B free probability

Appropriate notion of free independence, free convolution ⊞B , etc.


Appropriate notion of free independence, free convolution ⊞B , etc.Connection with infinitesimal probability (A, φ, φ′): two types oftype B non-crossing partitions: either no zero block (“φ part”) orhaving a zero block (“φ′ part”).


Appropriate notion of free independence, free convolution ⊞B , etc.Connection with infinitesimal probability (A, φ, φ′): two types oftype B non-crossing partitions: either no zero block (“φ part”) orhaving a zero block (“φ′ part”). There is a good analytical theory:

Theorem (Belinschi+DS ’12)

Let (µ1, µ′1) and (µ2, µ

′2) be infinitesimal laws: µj measures and µ′

j

distributions satisfying certain conditions. Let Xj(t) ∈ (A, φ) so

that µXj (t) ∼ µj + tµ′j + O(t2), and assume X1(t),X2(t) are free

for all t. Then Y (t) = X1(t) + X2(t) ∼ η + tη′ + O(t2) where:

◮ η = µ1 ⊞ µ2

◮ Gη′ = Gµ′

1(ω1(z))ω

′1(z) + Gµ′

2(ω2(z))ω

′2(z), where ωi are

subordination functions Gη = Gµi◦ ωi .


Appropriate notion of free independence, free convolution ⊞B , etc.Connection with infinitesimal probability (A, φ, φ′): two types oftype B non-crossing partitions: either no zero block (“φ part”) orhaving a zero block (“φ′ part”). There is a good analytical theory:

Theorem (Belinschi+DS ’12)

Let (µ1, µ′1) and (µ2, µ

′2) be infinitesimal laws: µj measures and µ′

j

distributions satisfying certain conditions. Let Xj(t) ∈ (A, φ) so

that µXj (t) ∼ µj + tµ′j + O(t2), and assume X1(t),X2(t) are free

for all t. Then Y (t) = X1(t) + X2(t) ∼ η + tη′ + O(t2) where:

◮ η = µ1 ⊞ µ2

◮ Gη′ = Gµ′

1(ω1(z))ω

′1(z) + Gµ′

2(ω2(z))ω

′2(z), where ωi are

subordination functions Gη = Gµi◦ ωi .

Can also consider multiplicative convolution ⊠B etc.

Asyptotic infinitesimal freeness

TheoremLet A(N) be a Gaussian random matrix and let B(N) be a

finite-rank matrix. Let τN be the joint law of A(N) and B(N) with

respect to N−1Tr . Then τN = τ + 1

Nτ ′ + o(N−1) and moreover

A(N) and B(N) are infinitesimally free under (τ, τ ′).

Asyptotic infinitesimal freeness

TheoremLet A(N) be a Gaussian random matrix and let B(N) be a

finite-rank matrix. Let τN be the joint law of A(N) and B(N) with

respect to N−1Tr . Then τN = τ + 1

Nτ ′ + o(N−1) and moreover

A(N) and B(N) are infinitesimally free under (τ, τ ′).

Corollary

Let A(N),B(N) ∈ (A, φ) be operators having the same law of

A(N) and B(N) respectively, but such that A(N) and B(N) are

free for each N. Then

µA(N)+B(N) = µA(N)+B(N) + o(1/N).

In particular,

µA(N)+B(N) = µA(N)⊞ µB(N) + o(1/N)

explaining the connection with free convolution.

Example.

Let BN = θE11 with E11 rank one projection with entry 1 inposition 1, 1 and zero elsewhere.Then

µBN = δ0 +1

N(δθ − δ0).

If AN is a Gaussian random matrix and η is the semicircle law, then

µAN = η + O(N−2).

So: µAN+BN = µ+ 1

Nµ̇+ o(1/N) and

(µ, µ̇) = (η, 0) ⊞B (δ0, δθ − δ0).

Example, ctd.

(µ, µ̇) = (η, 0) ⊞B (δ0, δθ − δ0)

Example, ctd.

(µ, µ̇) = (η, 0) ⊞B (δ0, δθ − δ0)

µ = semicircule law, Gµ(z) = z−√

z2 − 2, z ∈ C+∪(R\{±

√2}

Let

Gµ̇(z) =

ˆ

1

z − td µ̇(t)

Example, ctd.

(µ, µ̇) = (η, 0) ⊞B (δ0, δθ − δ0)


z2 − 2, z ∈ C+∪(R\{±

√2}

Let

Gµ̇(z) =

ˆ

1

z − td µ̇(t)

General theory implies:

Gµ̇(z) = ∂z

ˆ

log(z − t)d µ̇(t)

= ∂z

ˆ

1

z − t[h+(t)− h−(t)]dz , h± monotone

Example, ctd.

(µ, µ̇) = (η, 0) ⊞B (δ0, δθ − δ0)


z2 − 2, z ∈ C+∪(R\{±

√2}

Let

Gµ̇(z) =

ˆ

1

z − td µ̇(t)

General theory implies:

Gµ̇(z) = ∂z

ˆ

log(z − t)d µ̇(t)

= ∂z

ˆ

1

z − t[h+(t)− h−(t)]dz , h± monotone

= F ′µ(z)

(1

Fµ(z)− θ− 1

Fµ(z)

)

, Fµ(z) =1

Gµ(z).

Example, ctd.

Formula for ⊞B involving subordination functions gives:

Gµ̇(z) = F ′µ(z)

(1

Fµ(z)− θ− 1

Fµ(z)

)

= ∂z log

(Fµ(z)− θ

Fµ(z)

)

= ∂z log(1 − θGµ(z))

Example, ctd.


Gµ̇(z) = F ′µ(z)

(1

Fµ(z)− θ− 1

Fµ(z)

)

= ∂z log

(Fµ(z)− θ

Fµ(z)

)

= ∂z log(1 − θGµ(z)) = ∂z

ˆ

(z − t)−1[h+(t)− h−(t)]dt.

Example, ctd.


Gµ̇(z) = F ′µ(z)

(1

Fµ(z)− θ− 1

Fµ(z)

)

= ∂z log

(Fµ(z)− θ

Fµ(z)

)

= ∂z log(1 − θGµ(z)) = ∂z

ˆ

(z − t)−1[h+(t)− h−(t)]dt.

ˆ

[h+(t)− h−(t)]

z − tdt = log (1 − θGµ(z)) = log(1−θ(z−

√

z2 − 2)).

Example, ctd.

ˆ

[h+(t)− h−(t)]


√

z2 − 2))

Example, ctd.

ˆ

[h+(t)− h−(t)]


√

z2 − 2))

Recover dµ̇dt

= ∂t(h+(t)− h−(t)) by a kind of Stiletjes inversionformula.

Example, ctd.

ˆ

[h+(t)− h−(t)]


√

z2 − 2))

Recover dµ̇dt

= ∂t(h+(t)− h−(t)) by a kind of Stiletjes inversionformula.If θ > 1/

√2: let θ′ be the solution to Gη(θ

′) = 1/θ. Then

µ̇ = δθ′ −θ(t − 2θ)

π(2θ(t − θ)− 1)√

2 − t2χ[−

√2,√

2]dt

Example, ctd.

ˆ

[h+(t)− h−(t)]


√

z2 − 2))

Recover dµ̇dt



′) = 1/θ. Then

µ̇ = δθ′ −θ(t − 2θ)

π(2θ(t − θ)− 1)√

2 − t2χ[−

√2,√

2]dt

If θ < 1/√

2: no (real) solution to Gη(θ′) = 1/θ. Then

µ̇ =θ(t − 2θ)

π(2θ(t − θ)− 1)√

2 − t2χ[−

√2,√

2]dt.

Example, ctd.

ˆ

[h+(t)− h−(t)]


√

z2 − 2))

Recover dµ̇dt



′) = 1/θ. Then

µ̇ = δθ′ −θ(t − 2θ)

π(2θ(t − θ)− 1)√

2 − t2χ[−

√2,√

2]dt

If θ < 1/√

2: no (real) solution to Gη(θ′) = 1/θ. Then

µ̇ =θ(t − 2θ)

π(2θ(t − θ)− 1)√

2 − t2χ[−

√2,√

2]dt.

Thus dµN = 1

π

√2 − t2χ[−

√2,√

2] +1

Nµ̇+ O(N−2).

Numerical simulation

Average of 40 complex 100 × 100 matrices, with θ = 4 or θ = 0.4.

Ideas of proof

It turns out that µAN = µ+ O(1/N2). On the other hand, if Eij is

the matrix with 1 in the i , j-th entries and zeros elsewhere, then forany fixed p, 1

NTr(p({Eij}) = p(0) + 1

Nτ̇(p). For example, the law

of θE11 is δ0 +1

N(δθ − δ0).

LemmaThen for any polynomials q1, . . . , qr ,

limN→∞

ETr[

Eir j1q1(A(N))Ei1 j2q2(A(N))Ei2 j3 ×

· · · × Eir−1jrqr (A(N))]

=

r∏

s=1

δjs=is τ(qs) i.e.

limN→∞

ETr(Eir j1q1Ei1j2q2Ei2j3 · · · qr ) =r∏

s=1

limN

ETr(Ejs jsqsEis is )

Compute or use concentration.

Infinitesimal freeness is not for free!

Let Y(1)N , Y

(2)N be an N × N real iid self-adjoint Gaussian matrices.

Then each has law µN = η + 1

Nη′ + O(N−2) . However, they are

not asymptocially infinitesimally free.


Let Y(1)N , Y




not asymptocially infinitesimally free. Indeed,

Y(1)N ∼ 1√

K

K∑

j=1

Y(j)N


Let Y(1)N , Y





Y(1)N ∼ 1√

K

K∑

j=1

Y(j)N

and so if inf. freeness were to hold we would get by CLT

(η, η′) = (scaling by 1/√K ) ((η, η′)⊞B · · ·⊞B (η, η′))

︸︷︷︸

K

→ (ν, ν ′)

where (ν, ν ′) is an infinitesimal semicircle law (ν is semicircular,ν ′ = arcsine − semicircular).


Let Y(1)N , Y





Y(1)N ∼ 1√

K

K∑

j=1

Y(j)N

and so if inf. freeness were to hold we would get by CLT

(η, η′) = (scaling by 1/√K ) ((η, η′)⊞B · · ·⊞B (η, η′))

︸︷︷︸

K

→ (ν, ν ′)

where (ν, ν ′) is an infinitesimal semicircle law (ν is semicircular,ν ′ = arcsine − semicircular). But computation shows [Johannsen]that η′ = 1

4(δ√

2+ δ−

√2)− 1

2π1√

1−t2χ[−

√2,√

2]dt is not the arcsine

law.

Remarks

◮ Same statement holds if we assume thatA(N) = U(N)D(N)U(N)∗ with U(N) Haar-distributed unitarymatrix and D(N) a diagonal matrix so that µD

N are allsupported on a compact set and µD

N → µD weakly.

◮ Can also handle the real Gaussian case, which is different inthat µA

N = η + 1

Nη̇ + o(1/N) with η the semicircle law and

η̇ =1

4(δ√

2+ δ−

√2)− 1

2π√

2 − t2χ[−

√2,√

2](t)dt

˙µreal = µ̇complex + η̇.

◮ We can also deduce formulas for other polynomials in A(N)and B(N), such as products B(N)A(N)2B(N).

Thank you!

FREE PROBABILITY OF TYPE B AND ASYMPTOTICS OF …Many B’s around: bi-free probability...

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